Question about current draw for the RGB LEDs

[rezolution]

OK, maybe someone can explain this to me

Sooooooooo I slapped a meter on my VA Bane ('Resistorized Red' + 'Blue' ). I get 0.65 Amp with both dies active with Sanyo Eneloops and 0.75 Amp with Energizer E2 Lithiums with both dies active.

I got out my RGB Prophecy (R,G,B with 3 direct drive latching switches). I put a meter on that and i got:

Sanyo Eneloop 4.8V Nominal, 4.7V under one or two die load, 4.6V under three die load
Green die only = 0.50 Amp total battery draw
Blue die only = 0.50 Amp total battery draw
Red die only = 0.60 Amp total battery draw
Green+Blue dies = 0.60 Amp total battery draw
Red+Green dies = 0.70 Amp total battery draw
Red+Blue dies = 0.70 Amp total battery draw
Red+Green+Blue dies = 0.75 Amp total battery draw

I was told (by someone in the forums a while back) that when you turn two on, it uses twice the current of a single die and if you turned 3 on, it uses three times the current of a single die. It doesn't appear that way...

I'm really at a loss here...

[Ronin]

While I will concur in seeing the same post about 3 times the output when all switches are on I have to argue that it doesn't matter much, if anything it's better this way. The standard single switches draw the least battery power, and it gradually increases from those values, but in terms of brightness this only proves that any combination is brighter than a single switch. Plus it prolongs the battery life, making switching (or recharging) less of a repetitive hassle.

I'm just not sure what you're arguing, unless this is just an experiment, in which case good luck, mtfbwy.

[rezolution]

I'm not looking to argue.  I'm looking for someone to explain why they draw .5 by themselves but only .75 max when all three are on.  They are in parallel, so they should be drawing 0.5A+0.5A+0.6A = 1.6A

I don't understand why they only draw 0.75A when all three are on.  I would like someone who knows something about diodes to explain how the power consumption works for the way that Ultrasabers wires up the RGB saber setup on the RGB LEDs.

[vash1128]

Note: this is going to be boring unless you're curious about RGB LEDs or into electrical engineering stuff.

You could set it up in a few different ways, but I'm guessing the circuit is probably a single resistor wired in series with a set of 3 LEDs, which are wired in parallel.  The resistor is sized to match the forward voltage of the LEDs, which also sets the total current draw from the power supply.  This current is then (theoretically) distributed equally among the number of LEDs that are on.  To double the amount of current, you would have to change the resistance value.

I can think of two reasons you are getting variations in current draw (0.5A for one die, 0.75A for three die).  One, the forward voltage of LEDs is usually a range.  For example, the rebel star is min 2.55 and max 3.51.  Second, the green and blue rebel stars seem to have similar electrical characteristics, but the red seems to have a lower forward voltage.  This is sort of reflected in your data: green only and blue only are similar, red + green and red + blue are similar.

So in essence, what is happening when you turn on multiple dies is that your LED voltage drop and power consumption (P = V*I) are staying relatively constant (maybe going up a little), but the current is distributed between the LEDs.  This would cause them to get dimmer, but it's not a linear relationship.  For example, the green rebel star is 70 lumens @ 350mA, but 123lm @ 700mA.  Just because you halve the current doesn't mean the illumination halves.  So each LED is dimmer, but the mixed color should have a similar brightness to the single colors.  This is why your arctic blue is not twice as bright as a consular green, but maybe a little brighter.

There's a bit of speculation here, because I don't know the exact circuit diagram or the RGB LED type, but it seems to make sense to me from an EE standpoint.  Hopefully this satisfies your itch to understand how a RGB LED works!  Man this is like doing homework.

[rezolution]

Ah, this is EXACTLY what I was looking for.  Than k you very much!

I didn't know that the resistor was what was limiting the current.  That makes sense because the voltage of the dies couldn't be anywhere near 6v (or 4.8V if you're using rechargeables) so you would have to use a resistor to knock it down to 3ish volts.

I don't understand why the resistor limits the current though for the entire circuit.  I know you said that you'd have to change the resistor value to change the amount of power that can flow through the circuit but did you mean to change the resistor value in OHMs or to change the power the resistor dissipates?

For instance, if it's a 3 watt resistor, it can only allow 3 watts total to flow through the circuit.  If you put a 6 watt resistor in, it would allow 6 watts total to flow through?

[vash1128]

You're welcome!  So the power rating on the resistor is the max power that it can dissipate.  Basically, it's how much heat conductive material the resistor has to dissipate that amount of power.  I've seen a 150W resistor, which is basically a thin wire wrapped around a cylinder of ceramic, to dissipate the heat.  The ohms is the resistance value, which is connected to the size and material of the wire itself.  A larger diameter wire has smaller resistance, and certain materials, like copper, have better conductive properties and thus lower resistance.

It's important to get the resistance (in ohms) correct when sizing circuits so that Ohm's law (V = I*R) comes out correctly.  The power rating is there to ensure that the resistor doesn't pop due to the heat.  To change the amount of current flowing through the circuit, you would have to change the resistance, which is measured in ohms.  According to V=IR, lowering the resistance would raise the current for a given power supply voltage and LED forward voltage.  In this circuit, the resistor is the sole impedence, so it is the only "R" term, thus the sole determiner of current flow.

[rezolution]

So you decide how much current an LED gets by putting a resistor in series with it in a circuit?  If for instance you wanted the LED to get 1A of current (you have a 6V source and a 3V forward voltage on the LED), you would use a 3 Ohm Resistor in a 6V circuit? 3V = 1 Amp x 3 Ohm  (3 Watt Resistor needed)

If you wanted the LED to get 500mA of current to it, you would use a  6 Ohm resistor? 3V = .5 A x 6 Ohm (1.5 Watt Resistor needed)

What happens if you don't have a resistor in circuit?  Does the LED just die in a few seconds?

[vash1128]

First of all, current is measured in amps, and power (current*voltage) is measured in watts.  Let's not mix up our units.  Power can also be measured in horsepower, so you can talk about your 0.004 horsepower LED .

Yes, I think your calculations are correct.  Remember, you're applying V=IR to the resistor, so the V in the equation is the voltage drop across the resistor, i.e. Vsupply - Vled.  I can't tell if the 3V in your equation is the LED forward voltage or 6V - 3V. 

If you don't have a resistor, then I think it would be like a short circuit, which is very dangerous.  You would be applying 6V to your 3V LED with infinite current, and something, if not everything, would blow up.  The wires have a small amount of resistance in them, but still, it's not something I would try.

[rezolution]

Ok, I edited my post from power to current.  LoL

Thanks for all of your help.  It's greatly appreciated!

[vash1128]

Sure it was my pleasure!  Let me know if you have any more questions and mtfbwy!

[ThreadJack]

I don't speak your techie language.